Question

Energy of the Nucleus. How much work is needed to assemble an atomic nucleus containing three protons (such as Li) if we model it as an equilateral triangle of side$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}$with a proton at each vertex? Assume the protons started from very far away.

Short answer

The amount of work done for assemble three protons is2.16 MeV

Step-by-step solution

Determine the formula for the work done to assemble the three charges.

Consider the expression for the work done:

$\mathbf{U}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{0}}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{}}\frac{{\mathbf{q}}_{\mathbf{i}}}{{\mathbf{r}}_{\mathbf{i}}}$

Here,$q_0=+e$for three protons and r will be same for three.

Rewrite the expression as:

$$\begin{gathered} U=\left(\frac{+e}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{+3e}{r}\right) \\ U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{+3e^2}{r}\right) \end{gathered}$$ ….. (1)

Determine the work done for assemble three protons

Substitute $9.0\times {10}^9\frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}$for $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$,$1.6\times {10}^{-19}\mathrm{C}$for e, $2.00\times {10}^{-15}\mathrm{m}$for rin the equation (1) to determine the word done.

$$\begin{gathered} U=\left(9.0\times {10}^9\frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}\right)\left(\frac{3\left(1.6\times {10}^{-19}\mathrm{C}\right)}{\left(2.00\times {10}^{-15}\mathrm{m}\right)}\right) \\ \mathrm{U}=3.46\times {10}^{-13}J \\ U=\frac{3.46\times {10}^{-13}\mathrm{J}}{1.6\times {10}^{-19}\mathrm{C}} \\ U=2.16\mathrm{MeV} \end{gathered}$$

Hence, the amount of work done for assemble three protons is 2.16 MeV